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The question was asking me to find an expression for the constructive interference between light which follows the paths $PQ$ and $PRS$, once the light has reached the air outside of the two glass plates.

I found the correct expression:

$$2 \pi \left ( \frac {2d}{ \lambda \cos \theta} - \frac {2d \sin \theta \tan \theta} { \lambda} \right) + \pi = 2 \pi$$

But I don't understand why the expression is the phase change over path $PRS$ - phase change over path $PQ = 2 \pi $ and not just the phase change over path $PRS = 2 \pi $.

Can someone explain to me why I must take the path $PQ$ into account?

  • $\begingroup$ The difference does not have to be $2\pi$, but could actually be any integer multiple of $2\pi$. $\endgroup$ Commented Aug 30, 2017 at 17:46
  • $\begingroup$ Note that $\frac {1}{\cos \theta} - \sin \theta \tan \theta = cos \theta$ which will simplify your expression. $\endgroup$
    – Farcher
    Commented Aug 30, 2017 at 18:39

2 Answers 2


S and P need to be in phase, because light from S and Q reaches a distant point at the same time (with the help of a lens, if you like). This is because QS is a putative wavefront, being at right angles to the propagation direction.

[Mistake edited out.]

  • $\begingroup$ The second term was actually ((2dsinθtanθ)/n)/(λ/n) but the n's cancel. $\endgroup$ Commented Aug 31, 2017 at 8:43
  • $\begingroup$ Don't you mean S and Q need to be in phase? $\endgroup$
    – nasu
    Commented Sep 1, 2017 at 15:55
  • $\begingroup$ @nasu. I do. Mistake corrected. All in all a pretty disastrous attempt at an answer! $\endgroup$ Commented Sep 2, 2017 at 10:18

At P the incoming waves go along one of two paths, along PQ and along PRS. At point P whichever way the waves go they are in phase with one another.

The waves then are made to overlap and so you get interference.

If you took a snapshot of he wave you would find that the phase of the wave at Q would not be the same as the phase of the wave at P.

Likewise the phase of the wave at S would not be the same as the phase of the wave at P.

What you are interested in is the phase difference between Q and S.

This phase difference is give by the phase difference between S and P minus the phase difference between Q and P.
That phase difference is related to the difference in the optical path lengths that the waves have travelled.

  • $\begingroup$ Can you correct my reasoning here, what I was thinking was that if the phase change over PRS was equal to 2π then the light at S would be in phase with the light at point P. And so the two rays of light would be in phase, thus giving constructive interference. $\endgroup$ Commented Aug 30, 2017 at 16:53
  • 1
    $\begingroup$ You are correct but you are observing the phase difference which has left P and gone along two different paths so the length of each of those paths is important. $\endgroup$
    – Farcher
    Commented Aug 30, 2017 at 18:34
  • $\begingroup$ Maybe what @DT770 is missing is that the wave propagates in a direction perpendicular to the wavefront. The final direction of propagation is in the direction PQ, so the wavefront needs to be along QS, which is perpendicular to PQ. The phase of the wave is the same all along the wavefront, so it must be the same at P and at Q. $\endgroup$
    – S. McGrew
    Commented Oct 27, 2018 at 0:32

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